F.4 Maths probability(2)

Question 1
The probability that Henry takes one draw only
= Pr(Henry does not need the 2nd draw which is the 3rd draw in total)
= Pr(The drawing will finish before the 3rd draw)
= Pr(the final draw is the 1st draw) + Pr(the final draw is the 2nd draw)
= Pr(Henry gets the blue ball in the 1st draw) + Pr(John gets the blue ball in the 2nd draw which is his 1st draw)
= Pr(Henry gets the blue ball in the 1st draw) + Pr(Henry cannot get the blue ball in the 1st draw but John gets the blue ball in the 2nd draw which is his 1st draw)
= 1/9 + (8/9)×(1/8)
= 1/9 + 1/9
= 2/9

Question 2
The understanding in English is very important this time.
What is the meaning of "4 cars are not parked next to each other"?

Do you regard
[X] [X] [ ] [X] [X] [ ] [ ]
as 4 cars parked next to each other?

Interpretation 1
The only situation for 4 cars not parked next to each other is this:
[X] [ ] [X] [ ] [X] [ ] [X]

That is, the occupied places must be the 1st, 3rd, 5th and 7th.

The required probability is
　₄P₄ / ₇P₄
= 4! / (7! / 3!)
= (4 × 3 × 2 × 1) / (7 × 6 × 5 × 4)
= 1/35

Interpretation 2
The situations for 4 cars parked next to each other include these cases:
[X] [X] [X] [X] [ ] [ ] [ ]
[ ] [X] [X] [X] [X] [ ] [ ]
[ ] [ ] [X] [X] [X] [X] [ ]
[ ] [ ] [ ] [X] [X] [X] [X]

Then, the situations for 4 cars not parked next to each other are those other than the above cases.

The required probability is
　1 - 4 × ₄P₄ / ₇P₄
= 1 - 4 × 4! / (7! / 3!)
= 1 - 4 × (4 × 3 × 2 × 1) / (7 × 6 × 5 × 4)
= 1 - 4 × 1/35
= 1 - 4/35
= 31/35

2014-06-15 08:26:44 補充：
Question 1 補充回答：

題目問亨利只抽一次的機會。

留意抽球的次序：
（１）亨利　（２）約翰　（３）亨利　（４）約翰　．．．

題目要求的情況是亨利只抽一次，即是抽球活動於（１）或（２）之後結束。
這兩個情況也可使亨利不能抽第二次。

你忽略了的是第二個情況，即（１）亨利抽不到藍球，但（２）約翰抽到藍球。
這情況也符合題目要求，即亨利不能抽第二次，亦即只抽一次。

明白嗎？